1) Pareto charts are used to: A) identify inspection points in a process. B) outline production schedules. C) organize errors, p
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Answer:
the net work per cycle Btu per cycle
the power developed by the engine, W = 88.0144746 hp
Explanation:
the information given includes;
diameter of the four-cylinder bore = 3.7 in
length of the stroke = 3.4 in
The clearance volume = 16% = 0.16
The cylindrical volume
the crankshaft N rotates at a speed of 2400 RPM.
At the beginning of the compression , temperature = 60 F = 519.67 R
and;
Otto cycle with a pressure = 14.5 lbf/in² = (14.5 × 144 ) lb/ft²
= 2088 lb/ft²
The maximum temperature in the cycle is 5200 R
From the given information; the change in volume is:
the mass in air ( lb) can be determined by using the formula:
where;
R = 53.3533 ft.lbf/lb.R°
m = 0.0018962 lb
From the tables of ideal gas properties at Temperature 519.67 R
At state of volume 2; the relative volume can be determined as:
The specific energy at
is 184.7 Btu/lb
From the tables of ideal gas properties at maximum Temperature T = 5200 R
To determine the relative volume at state 4; we have:
The specific energy at
is 591.84 Btu/lb
Now; the net work per cycle can now be calculated as by using the following formula:
Btu per cycle
the power developed by the engine, in horsepower. can be calculated as follows;
In the four-cylinder, four-stroke internal combustion engine; the power developed by the engine can be calculated by using the expression:
where ;
N' = 1200 cycles/min
N' = 1200 cycles/60 seconds
N' = 20 cycles/sec
W = 4 × 20 cycles/sec × 0.777593696
W = 62.20749568 Btu/s
W = 88.0144746 hp
Answer:
flow(m) = 54.45 kg/s
thermal efficiency u = 44.48%
Explanation:
Given:
- P_1 = P_8 = 10 KPa
- P_2 = P_3 = P_6 = P_7 = 800 KPa
- P_4 = P_5 = 10,000 KPa
- T_5 = 550 C
- T_7 = 500 C
- Power Output P = 80 MW
Find:
- The mass flow rate of steam through the boiler
- The thermal efficiency of the cycle.
Solution:
State 1:
P_1 = 10 KPa , saturated liquid
h_1 = 192 KJ/kg
v_1 = 0.00101 m^3 / kg
State 2:
P_2 = 800 KPa , constant volume process work done:
h_2 = h_1 + v_1 * ( P_2 - P_1)
h_2 = 192 + 0.00101*(790) = 192.80 KJ/kg
State 3:
P_3 = 800 KPa , saturated liquid
h_3 = 721 KJ/kg
v_3 = 0.00111 m^3 / kg
State 4:
P_4 = 10,000 KPa , constant volume process work done:
h_4 = h_3 + v_3 * ( P_4 - P_3)
h_4 = 721 + 0.00111*(9200) = 731.21 KJ/kg
State 5:
P_5 = 10,000 KPa , T_5 = 550 C
h_5 = 3500 KJ/kg
s_5 = 6.760 KJ/kgK
State 6:
P_6 = 800 KPa , s_5 = s_6 = 6.760 KJ/kgK
h_6 = 2810 KJ/kg
State 7:
P_7 = 800 KPa , T_7 = 500 C
h_7 = 3480 KJ/kg
s_7 = 7.870 KJ/kgK
State 8:
P_8 = 10 KPa , s_8 = s_7 = 7.870 KJ/kgK
h_8 = 2490 KJ/kg
- Fraction of steam y = flow(m_6 / m_3).
- Use energy balance of steam bleed and cold feed-water:
E_6 + E_2 = E_3
flow(m_6)*h_6 + flow(m_2)*h_3 = flow(m_3)*h_3
y*h_6 + (1-y)*h_3 = h_3
y*2810 + (1-y)*192.8 = 721
Compute y: y = 0.2018
- Heat produced by the boiler q_b:
q_b = h_5 - h_4 +(1-y)*(h_7 - h_8)
q_b = 3500 -731.21 + ( 1 - 0.2018)*(3480 - 2810)
Compute q_b: q_b = 3303.58 KJ/ kg
-Heat dissipated by the condenser q_c:
q_c = (1-y)*(h_8 - h_1)
q_c= ( 1 + 0.2018)*(2810 - 192)
Compute q_c: q_c = 1834.26 KJ/ kg
- Net power output w_net:
w_net = q_b - q_c
w_net = 3303.58 - 1834.26
w_net = 1469.32 KJ/kg
- Given out put P = 80,000 KW
flow(m) = P / w_net
compute flow(m) flow(m) = 80,000 /1469.32 = 54.45 kg/s
- Thermal efficiency u:
u = 1 - (q_c / q_b)
u = 1 - (1834.26/3303.58)
u = 44.48 %
